Problem: Simplify the following expression: $y = \dfrac{4x^2+7x+3}{4x + 3}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(4)}{(3)} &=& 12 \\ {a} + {b} &=& &=& {7} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $12$ and add them together. The factors that add up to ${7}$ will be your ${a}$ and ${b}$ When ${a}$ is ${3}$ and ${b}$ is ${4}$ $ \begin{eqnarray} {ab} &=& ({3})({4}) &=& 12 \\ {a} + {b} &=& {3} + {4} &=& 7 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({4}x^2 +{3}x) + ({4}x +{3}) $ Factor out the common factors: $ x(4x + 3) + 1(4x + 3)$ Now factor out $(4x + 3)$ $ (4x + 3)(x + 1)$ The original expression can therefore be written: $ \dfrac{(4x + 3)(x + 1)}{4x + 3}$ We are dividing by $4x + 3$ , so $4x + 3 \neq 0$ Therefore, $x \neq -\frac{3}{4}$ This leaves us with $x + 1; x \neq -\frac{3}{4}$.